Item 1.
All we need to do here is divide both sides by 7. We add 13 to the right-hand side repeatedly until we get a multiple of 7 (adding 13 is the same as adding 0, so this is legal). We get \(25\text{,}\) \(38\text{,}\) \(51\text{,}\) \(64\text{,}\) \(77\) – got it. So we have:
\begin{equation*}
\begin{aligned}7x \amp \equiv 12 \pmod{13} \\
7x \amp \equiv 77 \pmod{13} \\
x \amp \equiv 11 \pmod{13}.
\end{aligned}
\end{equation*}